#### Answer

The proof is detailed in the Work step by step below.

#### Work Step by Step

$$y=mx+b$$
At any point $(x_0,mx_0+b)$, we have $$y'=\lim_{h\to0}\frac{m(x_0+h)+b-(mx_0+b)}{h}$$
$$y'=\lim_{h\to0}\frac{mx_0+mh+b-mx_0-b}{h}=\lim_{h\to0}\frac{mh}{h}$$
$$y'=\lim_{h\to0}m=m$$
So the tangent line at any point $x=x_0$ will have this form: $$y=mx+n$$
To find $n$, we use the coordinates of the given point $(x_0,mx_0+b)$:
$$mx_0+n=mx_0+b$$
$$n=b$$
Therefore, the tangent line of the given line is $y=mx+b$, which is exactly the same line.
This means the line $y=mx+b$ is its own tangent line at any point $(x_0,mx_0+b)$.